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Math Help - Bouncing Balls

  1. #1
    Member classicstrings's Avatar
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    Bouncing Balls

    Balls are manufactured according to the model y = 5mod(e^(-at)sin3t).

    Balls are considered flat if they fail to rebound to a height of 3m. Find the greatest value of a for a ball to be considered flat.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Balls are manufactured according to the model y = 5mod(e^(-at)sin3t).

    Balls are considered flat if they fail to rebound to a height of 3m. Find the greatest value of a for a ball to be considered flat.
    I presume that y is the height the ball bounces to?

    y = 5*|e^{-at}*sin(3t)|

    We want to find the a value such that y(max) = 3. What then, is the maximum height of the ball for variable a?

    Of course you can do this with Calculus, but just look at the function y(t) for a moment. The e^{-at} is going to modify the sine function such that the magnitude of the max and min y values are going to be reduced as time goes on. (ie. The envelope of the sine function decays exponentially, if you understand the terminology.) So the largest maxima value for y will be at the first maxima value for sin(3t), which is at t = (pi)/6 s. So this is the t value for the maximum value of y.

    Thus
    y(max) = 5*|e^{-a*(pi)/6}*sin(3*(pi)/6)| = 5*e^{-a*(pi)/6}

    which we need to set to 3 m for the largest possible a value. So solve:
    3 = 5*e^{-a*(pi)/6}

    e^{-a*(pi)/6} = 3/5

    -a*(pi)/6 = ln(3/5) = ln(3) - ln(5)

    a*(pi)/6 = ln(5) - ln(3)

    a = 6/(pi) * [ln(5) - ln(3)] = 0.267468 (or so)

    Any larger a value will produce a first maximum in y that is less than 3 m. (You can also check this with Calculus, but you should be able to use simple logic to see this.)

    -Dan
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    Member classicstrings's Avatar
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    Sorry if I didnt make it clear, the "rebound height" is the second maximum of the graph where t >0.

    How would it be worked out using derivatives?

    I also need to work out the exact value of t that gives the time when the ball rebounds to the first rebound height i.e. the second maximum. And comment after sketching a graph of t vs a on what happens as a increases to the first rebound point.
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    Edit wrong
    Last edited by classicstrings; May 9th 2007 at 03:07 PM. Reason: mistake
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Sorry if I didnt make it clear, the "rebound height" is the second maximum of the graph where t >0.

    How would it be worked out using derivatives?

    I also need to work out the exact value of t that gives the time when the ball rebounds to the first rebound height i.e. the second maximum. And comment after sketching a graph of t vs a on what happens as a increases to the first rebound point.
    Well the second maximum will be at:
    t = 3*(pi)/6 s = (pi)/2 s
    I'll let you work it from there.

    The Calculus version has us taking the time derivative of the y(t) function and solving for what times it is equal to 0.
    y(t) = 5*|e^{-at}*sin(3t)|

    y'(t) = sign(sin(3t))*15*e^{-at}cos(3t) - 5ae^{-at}|sin(3t)| = 0

    Solve that for t and you are in business. (I'm not about to do it!)

    -Dan
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